Distance between two points 

The distance between two points P(x1, y1) and Q(x2, y2) is (see the figure given below).

 

198_Distance between two points 1.JPG

 Length PQ = √(x2 – x1)2 + (y2 – y1)2

 Proof:

Let P(x1, y1) and Q(x2, y2) be the two points and let the distance between them be d. Draw PA, QR parallel to y-axis and PR parallel to x-axis. 

 

 

 Angle QRP = 90o  
 
                 ⇒ d2 = PR2 + RQ2 

 

                ⇒ d2 = (x2 – x1)2 + (y2 – y1)2

                ⇒ d = √(x2 – x1)2 + (y2 – y1)2. 

 

Section Formula   

Let us say we want to know the co-ordinates of point which divides a line segment between two points A(x1, y1) and B(x2, y2) in the ratio m : n. 

The coordinates of such a point are given by 

(nx1 + mx2/m+n, ny1 + my2/m+n) (for internal division)   

 
Note: 

This is called section formula. 

Let P divide the line segment AB in the ratio m : n. If P is inside AB then it is called internal division; if it is outside AB then it is called external division. 

However in each case AP/BP [or AP'/BP' or AP"/BP"] = m/n.  

 
Proof: 

        Consider ?   ABB’  

        Since BB’||PQ and AP:PB = m:n (see figure given below)

 

2370_Distance between two points 2.JPG

 

                AQ/AB' = PQ/BB' = m/m+n (= AP/AB) 

               ⇒ x – x1/x2 – x1 = m/m+n 

               ⇒ x = nx1 + mx2/m+n and y = ny1 + my2/m+n 

If P is outside AB (less assume it is at P’) 

We have 

                x – x1/x2 – x1 = m/m+n 

               ⇒ x = nx1 + mx2/m+n and y = ny1 + my2/m+n 

Similarly if P is at P” then 

                 x = –mx2+m+n/n–m, y = – my2+ny1/m+n

 

 

 Note:  

m:n can be written as m/n or λ:1. So any point on line joining A and B will be P(λx2+x1/λ+1.λy2+y1/λ+1). It is useful to assume λ:1 because it involves only one variable. 

  
illustration: 

Find the ratio in which line segment A(2, –1) and B(5, 2) is divided by x-axis. 

Solution:  

Let x-axis intersect line at point P(xp, 0) such that AP/BP = λ/1
⇒ yP = 0 = λy2+1.ya/λ+1 = 2λ+(–1)/λ+1 Þ λ = 1/2
⇒ AP/BO = 1/2

 
Illustration: 

Prove that altitudes of a triangle are concurrent and prove that the co-ordinates of the point of con-currency are 

(x1 tan A + x2 tan B + x3 tan C/tan A + tan B + tan C, y1 tan A + y2 tan B + y3 tan C/tan A + tan B + tan C), 

 
Solution: 

In triangle A(x1, y1), B(x2, y2) and C(x3, y3), draw AD perpendicular to BC. Our effort now should be to find the co-ordinates of the point D.

 

178_Distance between two points 3.JPG

 To do that, we need to find BC/CD. (figure is given above) 

                tan B = AD/BD and tan C = AD/CD 

            ⇒ BD/DC = tan C/tan B
 
Now we apply section formulae. 

        xD = x2 tan B + x3 tan C/tan B + tan C                                        …… (i)

            yD = y2 tan B + y3 tan C/tan B + tan C                                          …… (ii)

 We know that orthocenter will lie on AD. We need to find this point and its co-ordinates. 

We should select a point H1 on AD and take the ratio AH1/H1D in such a manner so that xH1 and yH1 calculated form (i) should be symmetric in x1, x2, x3, tan A, tan B and tan C. Think before you proceed 

Let    AH1/H1D = tan B + tan C/tan A 

                         ⇒ xH1 = x1 tan A + x2 tan B + x3 tan C/tan A + tan B + tan C

        and   ⇒ yH1 = y1 tan A + y2 tan B + y3 tan C/tan A + tan B + tan C 

Since the result is symmetric, this point H1 will lie on other altitude as well i.e. the altitudes are concurrent

 

                         ⇒ xH = xH1 and yH = yH1 

 
Illustration: 

Prove analytically that in a right angled triangle the midpoint of the hypotenuse is equidistant from the three angular points. 

 Solution: 

While proving a problem analytically take most convenient co-ordinates of known points. 

In the present case triangle is assumed as AOB with coordinates as shown in figure given below, C is midpoint of AB.

 

236_Distance between two points 4.JPG

 So co-ordinates of C will be (a/2, b/2) 

Now AB = √a2 + b2
 CA = CB = AB/2 (C is mid point of AB) 

 

        = √a2 + b2 

and, we know that the distance between two points C and O is given by 

CO = √(a/2 – 0)2 + (b/2 – 0)2 = √a2 + b2/2 

Hence CA = CB = CO 

Coordinates of the point P dividing the join of two points A(x1, y1) and

B(x2, y2) internally in the given ratio λ1 : λ2 i.e., AP/BP = λ1/λ2 areP(λ2x1+λ1x2/λ2+λ1λ2y1+λ1y2/λ2+λ1). 

Coordinates of the point P dividing the join of two points A(x1, y1) and B(x2, y2) externally in the ratio λ1 : λ2 i.e., Ao/BP = λ1/λ2 areP(λ2x1+λ1x2/λ2λ1λ2y1+λ1y2/λ2λ1).

 
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